Thrust Calculation: 1000 psi Tank, 1 in² Removed

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Discussion Overview

The discussion revolves around the calculation of thrust generated by a tank under pressure, specifically focusing on the implications of removing a square inch from the tank's end and how this affects the force exerted in the opposite direction. The conversation includes theoretical considerations, practical testing scenarios, and comparisons between using air and water as the working fluid.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Experimental/applied

Main Points Raised

  • Some participants propose that removing a square inch from a tank with 1000 psi creates a pressure differential that results in a 1000 lb force pushing in the opposite direction.
  • Others argue that according to Newton's Third Law, if there is a 1000 lb force on the removed area, there must be an equal reaction force on the tank.
  • A participant draws a parallel between the thrust generation in rockets and the pressure dynamics in the tank, suggesting that the combustion chamber acts similarly to a high-pressure tank.
  • There is a discussion about the practical differences between using air and water in the tank, with some noting that maintaining pressure with air requires moving a larger volume compared to water.
  • One participant shares details about their testing setup, including the pressures and equipment used, and expresses curiosity about expected thrust values in different scenarios.
  • Another participant highlights the importance of kinetic energy in the context of thrust generation, suggesting that compressed air may be more suitable for applications requiring energy storage.
  • There is a question regarding the direction of thrust and its relevance to the intended application, indicating that the design may influence the choice of fluid used.

Areas of Agreement / Disagreement

Participants express differing views on the implications of pressure changes in the tank and the resulting thrust. While some agree on the basic principles of thrust generation, there is no consensus on the specifics of the calculations or the practical outcomes of the tests being discussed.

Contextual Notes

Participants mention various assumptions, such as the need for constant pressure maintenance and the differences in energy storage between compressed air and liquids. The discussion also reflects uncertainty regarding the actual thrust values that may be achieved in practice.

bluechipx
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Let's say we have a tank with 1000 psi pushing in all directions, if we remove one square inch from an end, now the tank has 1000 psi more at one end than the other. Does this mean that there is a 1000 lb push in the opposite direction of the removed square inch?
 
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bluechipx said:
Let's say we have a tank with 1000 psi pushing in all directions, if we remove one square inch from an end, now the tank has 1000 psi more at one end than the other. Does this mean that there is a 1000 lb push in the opposite direction of the removed square inch?
If the force on that square inch jet of water is indeed 1000pounds force then Newton's Third law tells you that there is 1000pounds reaction force on the tank. The forces must be equal.
PS IS someone pumping water into the tank to keep the internal pressure constant?
 
bluechipx said:
Let's say we have a tank with 1000 psi pushing in all directions, if we remove one square inch from an end, now the tank has 1000 psi more at one end than the other. Does this mean that there is a 1000 lb push in the opposite direction of the removed square inch?
Yes, and that's one way of explaining how a rocket generates thrust in a vacuum where "there's nothing to push against". The combustion chamber of a rocket is basically a high-pressure tank with an opening at the back.
 
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sophiecentaur said:
If the force on that square inch jet of water is indeed 1000pounds force then Newton's Third law tells you that there is 1000pounds reaction force on the tank. The forces must be equal.
PS IS someone pumping water into the tank to keep the internal pressure constant?

I am making a test fixture that will monitor thrust which is still under construction. I will test both ways, just air and air pushing water. I'll try to post pics soon. The tank is a full size high pressure welding type cylinder rated for 4500 psi with a test pressure of 6750 psi. The ball valve is a two inch with a pressure rating of 5000 psi. With no measuring system, I started out with a mere 250 psi yesterday for the first test, then tried 600 next. I am focusing on snapping the valve open as quickly as possible for the best effect, which was very impressive for the low starting pressure.
 
bluechipx said:
, just air and air pushing water
There will be a big practical difference between to two situations. You will need to shift a lot of air, fast, to maintain the sort of pressure compared with doing it with water. But there is a lot of Potential Energy stored in compressed air, compared with the Energy stored in a compressed liquid.
Perhaps you could go into some more detail to help us make the leap between your actual problem and the original question. Was that the 'right' question?
 
Although the actual test will be what matters, I was just curious if there was an accurate way of knowing beforehand what to expect. I have a hydraulic fixture with a one square inch piston (1.128") that applies pressure to a 10,000 psi gauge. The gauge has a telltale pointer that records max pressure. The tank will be pressurized to 4500 psi and released instantly though a two inch dia ball valve. I will be testing both ways, air over water and air only. Anybody care to venture a guess as to the actual thrust both ways? The valve has an area of just over three square inches (3.14). Hopefully I may see 3.14X 4500 = 14,130 lbs, but I suspect in reality the figure will be much lower for reasons I am unaware of.
 
Which direction of thrust is important for you? Is this for a 'projectile' or a ram of some kind?
There is an important factor which I hinted at earlier. If you actually want some Kinetic Energy out of this device then compressed air is probably the way to go because it 'stores' energy whereas a compressed liquid stores virtually none. Otoh, the lack of stored energy makes things more controllable, if you simply want a ram. It's no coincidence that hydraulics are used for working tools and vehicles.
You mentioned a combination of air and water. That could have its advantages in some specialised applications.
Give us a clue.
bluechipx said:
Anybody care to venture a guess as to the actual thrust both ways?
The initial thrust will be the same as the pressure times the area but things will change once the fluid starts to flow.
 

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